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That's not the right way to count!

door x - winning, losing, losing
door y - losing, winning, losing
door z - losing, losing, winning

Clearly each door has a 1/3 chance of winning.

However the chance that either door y or door z is the winner and is randomly selected to be revealed is only 2/6 or 1/3. But that's not what we are interested in since only one of the two can be selected, the odds are only 1/2 of that or 1/6 as I stated above that the winning door will be revealled.

The situation is very clean.

Not by your table which assumes equal probabilities of outcomes. Your numbers would be correct if the random selection process included revealing door #1 but it doesn't and this needs to be accounted for.

I hope I explained this in a clear enough fashion.



Give me rhythm or give me death!

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